How to Multiply Fractions with Variables

When multiplying fractions, I am a fan of factorizing and canceling common factors instead of multiplying and then simplifying.

The reason for this is that it makes the calculations much easier, and I like it when things are easy. By cross-canceling when possible instead of expanding the parentheses, the calculations become simpler and less messy.

Here you will learn to multiply fractions with variables by factorizing and cross-canceling. Beware that you can only cancel factors that are on both sides of the fraction bar.

Rule

Multiplication Through Canceling

1.
Factorize the numerators.
2.
Factorize the denominators.
3.
Cancel common factors.
4.
Find the answer.

Example 1

Evaluate x + 1 x + 2 2x + 4 3x + 3

You factorize and cross-cancel in this way:

x + 1 x + 2 2x + 4 3x + 3 = (x + 1) (x + 2) 2 (x + 2) 3 (x + 1) = (x + 1) (x + 2) 2(x + 2) 3(x + 1) = 2 3

Example 2

Evaluate 2x + 6 x + 3 x2 9 x + 3

You factorize using the third algebraic identity of quadratic expressions, and then you cancel and multiply like this:

2x + 6 x + 3 x2 9 x + 3 = 2 (x + 3) (x + 3) (x + 3) (x 3) (x + 3) = 2(x + 3) (x + 3) (x + 3) (x 3) (x + 3) = 2 (x 3) = 2x 6

2x + 6 x + 3 x2 9 x + 3 = 2 (x + 3) (x + 3) (x + 3) (x 3) (x + 3) = 2(x + 3) (x + 3) (x + 3) (x 3) (x + 3) = 2 (x 3) = 2x 6

Example 3

Evaluate x2 1 x + 5 x2 25 x 1

Use the third algebraic identity of quadratic expressions to factorize, and then cross-cancel and multiply in this manner:

x2 1 x + 5 x2 25 x 1 = (x 1) (x + 1) (x + 5) (x + 5) (x 5) (x 1) = (x 1) (x + 1) (x + 5) (x + 5) (x 5) (x 1) = (x + 1) (x 5) = x2 5x + x 5 = x2 4x 5

x2 1 x + 5 x2 25 x 1 = (x 1) (x + 1) (x + 5) (x + 5) (x 5) (x 1) = (x 1) (x + 1) (x + 5) (x + 5) (x 5) (x 1) = (x + 1) (x 5) = x2 5x + x 5 = x2 4x 5

Example 4

Evaluate 2x2 32 x 4 x + 5 2 (x2 25)

You factorize with the help of the third algebraic identity of quadratic expressions, and then you cancel and multiply the fractions in this way:

= 2x2 32 x 4 x + 5 2 (x2 25) = 2 (x2 16) (x 4) (x + 5) 2 (x + 5) (x 5) = 2 (x + 4) (x 4) (x 4) (x + 5) 2(x + 5) (x 5) = (x + 4) 1 (x 5) = x + 4 x 5

2x2 32 x 4 x + 5 2 (x2 25) = 2 (x2 16) (x 4) (x + 5) 2 (x + 5) (x 5) = 2 (x + 4) (x 4) (x 4) (x + 5) 2(x + 5) (x 5) = (x + 4) 1 (x 5) = x + 4 x 5

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